A triangle has sides A, B, and C. The angle between sides A and B is pi/4 and the angle between sides B and C is pi/12. If side B has a length of 5, what is the area of the triangle?

2 Answers
Jun 14, 2017

The area is =2.64u^2

Explanation:

The angle between A and B is

theta=pi-(1/12pi+1/4pi)=pi-4/12pi=2/3pi

sintheta=sin(2/3pi)=sqrt3/2

We apply the sine rule to find =A

A/sin(1/12pi)=B/sin(2/3pi)

A=Bsin(1/12pi)/sin(2/3pi)

The area of the triangle is

a=1/2ABsin(1/4pi)

a=1/2Bsin(1/12pi)/sin(2/3pi)*B*sin(1/4pi)

=1/2*5*5*sin(1/12pi)*sin(pi/4)/sin(2/3pi)

=2.64

Jun 14, 2017

2.64 unit^2

Explanation:

Let say the angle, a = pi/12 and c = pi/4. Then angle between C and A, b = pi - 1/4 pi - 1/12 pi = 8/12 pi = 2/3 pi.

Area of triangle, = 1/2 BC sin a
= 1/2 (5)(C) sin (pi/12).->i

use, B/sin b = C/sin c to find C

5/(sin (2/3 pi)) = C /(sin (1/4 pi))

C =5/(sin (2/3 pi)) * sin (1/4 pi) = 4.08unit

Area of triangle, plug in C in ->i
= 1/2 (5)(4.08) sin (pi/12) = 2.64 unit^2